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U(s) is a continuous operator on H0 - In view of (15.6.5) and (5.5.1), this con-
tinuity is equivalent to the following condition on the form g  gxo:

(U) For each s e A there exists a real number Ms ^ 0 such that

(15.6.9)                                 g(st,st)^Msg(t, i)
for all 16 A.


(15.6.10)    Let g be a positive Hilbertform on A, satisfying the condition (U),
and suppose that (with the notation of (15.6.8)) the prehilbert space A/n is
separable, so that (6.6.2) it can be identified with a dense vector subspace of a
Hilbert space H. Then for each seA the endomorphism n(i) H* n(sf) of A/n
extends to a continuous endomorphism x \-> V(s) - x 0/H, and s \-> V(s) is a repre-
sentation ofAinVL.

If n(f) = 7i(O then n(st) = n(st)') because n is a left ideal. Hence the
endomorphism of H0 under consideration is well-defined, and the definition
of the scalar product on A/n, together with (15.6.9), shows that this endomor-
phism is continuous (5.5.1). Hence the existence of the continuous operator
V(s) (5.5.4). Since n((ss')t) = Ti(s(s'i)\ we have V(ss')= V(s)V(s'). Also, by
virtue of (15.6.3)

(KO*)  71(01 *(0) = &**, O = &, *0

= *<X, 0 = (V(s) ' n(t') | TU(O) = (n(t) | V(s) '

which shows that V(s*) = (V(s))*. Finally, if A has a unit element e, then
evidently V(e) = 1H , and therefore jh-> V(s) is a representation of A in H.

It is useful to know when the representation s\-+ V(s) so defined is non-
degenerate (15.5.5). This is equivalent to the following condition on g:

(N)   The elements n(sf)form a total set in the prehilbert space A/n.
This condition is trivially satisfied when A has a unit element.

Under the conditions of (1 5.6.1 0), there does not in general exist an element
x0 e H such that g(s, t) = (V(s) - x0 \ V(f) - ;c0) (cf. Section 15.9, Problem 3).
However, such a vector always exists when A has a unit element e : we may
take x0 = n(e).n be reconstructed from the algebra structure of A and the form